Evaluate $\int \tan^6x\sec^3x \ \mathrm{d}x$ Integrate $$\int \tan^6x\sec^3x \ \mathrm{d}x$$
I tried to split integral to $$\tan^6x\sec^2x\sec x$$ but no luck for me. Help thanks
 A: You can use the substitution $u = \sec + \tan$.
Note that $\dfrac 1u = \sec - \tan$, because $(\sec + \tan)(\sec - \tan) = \sec^2 - \tan^2 = 1$.
We have:
$$u^2 + 1 = [(\sec + \tan)^2] + [1] \\
= [\sec^2 + 2 \sec \tan + \tan^2] + [\sec^2 - \tan^2]\\
= 2\sec^2 + 2 \sec \tan \\
= \sec[2(\sec + \tan)] \\
= \sec(2u) \\
\implies \dfrac {(u^2 + 1)}{2u} = \sec
$$
Also:
$$
\begin{align}
u^2 - 1 &= [(\sec + \tan)^2] - [1]\\
&= [\sec^2 + 2 \sec \tan + \tan^2] - [\sec^2 - \tan^2]\\
&= 2\tan^2 + 2 \sec \tan\\
&= \tan[2(\sec + \tan)]\\
&= \tan(2u)\\
\end{align}
$$
$$\implies \dfrac {u^2 - 1}{2u} = \tan$$

Finally:
$$
\begin{align}
du &= (\sec + \tan)' dx \\ 
&= (\sec\tan + \sec^2) dx\\
&= \sec(\sec + \tan) dx\\
&= \dfrac {u^2 + 1}{2u}u dx\\
&= (u^2 + 1)\dfrac {dx}2\\
\end{align}
$$
$$\implies dx =\dfrac { 2du }{(u^2 + 1)}$$
So in our integral:
$$\sec \to \dfrac {u^2 + 1}{2u}\\
\tan \to \dfrac {u^2 - 1}{2u}\\
dx  \to \dfrac { 2}{(u^2 + 1)} du
$$
This technique gives us a series of powers of $u$
for any integrand $\tan^m\sec^{n+1}$.
A: Following your suggestion: 
From trig and algebra:
$$ \int \tan^6 x \sec^3 x\, dx   = \int \tan^6 x \sec^2 x\sec x\, dx  = \int \tan^6 x (1 + \tan^2 x) \sec x\, dx\\ = \int \tan^8 x \sec x \, dx+ \int \tan^6 x \sec x\,dx,  $$
so we are "interested" in evaluating
$$ I_n = \int \tan^{2n}x \sec x \, dx .$$ 
For $n>0$, use integration by parts (and the fact that the anti-derivative of $\tan x \sec x$  is $ \sec x $, and the derivative of $\tan x$ is $\sec^2 x $):
$$I_n = \int \tan^{2n -1 }x \tan x\sec x \, dx = \tan^{2n-1 }x \sec x - (2n -1 )\int \tan^{2n-2}x \sec^3 x \,dx.\tag{*}$$
Using trig again  on the last integral:
$$ \int \tan^{2n-2}x \sec^3 x \,dx = \int \tan^{2n-2}x \sec^2 x \sec x\,dx\\ = \int \tan^{2n-2}x (1 + \tan^2 x) \sec x\,dx = I_n + I_{n-1}.$$ 
Substitute into (*), and isolate $I_n$:
$$ I_n = 1/2n \left( \tan^{2n-1} x \sec x -(2n-1) I_{n-1} \right ).$$
Also, $$I_0 = \int \sec x \,dx = \ln ( \sec x + \tan x) + C .$$
So for the answer, curse and recurse, and combine all the above...
A: Here is what I got from Wolfram Alpha, condensed a little bit
$$\int tan^6(x)sec^3(x)dx$$
$$= \int (sec^2-1)^3sec^3(x)dx$$
$$= \int \Big(sec^9(x) -3sec^7(x)+3sec^5(x)-sec^3(x)\Big)$$
$$= \int sec^9(x)dx -3\int sec^7(x)dx+3\int sec^5(x)dx-\int sec^3(x)dx$$
Since $\int sec^m(x) = \frac{sin(x)sec^{m-1}(x)}{m-1} + \frac{m-2}{m-1}\int sec^{m-2}(x)dx$
$$\Rightarrow \frac{1}{8}tan(x)sec^7(x) -\frac{17}{18}\int sec^7(x)dx+3\int sec^5(x)dx-\int sec^3(x)dx$$
$$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{48}\int sec^5(x)dx-\int sec^3(x)dx$$
$$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{64}\int sec^3(x)dx$$
$$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{128}[tan(x)sec(x)-\int sec(x)dx]$$
$$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{128}tan(x)sec(x)-\frac{5}{128}log[tan(x)+sec(x)]$$
$$ = \frac{1}{384}\bigg(48tan(x)sec^7(x) -136tan(x)sec^5(x)+118tan(x)sec^3(x)-15tan(x)sec(x)-15log[tan(x)+sec(x)]\bigg)$$
A: You have
$$
\tan^6x\sec^3x=\frac{\sin^6x}{\cos^9x}=\frac{\sin^6x}{(1-\sin^2x)^5}\cos x
$$
so by substituting $t=\sin x$
$$
\int\frac{t^6}{(1-t^2)^5}dt
$$
that is long with partial fractions, but simple.
